# Taking "intermediate" limits, e.g. ${\cal X}\to 0$ but ${\cal X}/\epsilon \to \infty$ as $\epsilon\to 0$ - *without* taking $\epsilon \to 0$ directly

I have an expression that I would like to take limits of but not in a conventional sense. Take for example, an expression like This expression involves intermediate variables $${\cal X(\epsilon)},{\cal Y(\epsilon)}$$ whose limits are dependent on how they behave as $$\epsilon\to 0$$. I would like to take limits of the above in such a way that $${\cal X},{\cal Y}\to 0$$ but also that $${\cal X}/\epsilon,{\cal Y}/\epsilon\to\infty$$ as $$\epsilon\to 0$$. This will allow the exponential term to decay to zero, whilst any term involving only $${\cal X}$$ or $${\cal Y}$$ will go to zero. Strictly speaking, the first term inside the second pair of brackets, $${\cal Y}/\epsilon$$, diverges as $$\epsilon \to 0$$ but upon the action of multiplying by the prefactor $$\epsilon$$, it will decay to zero as $$\epsilon\to0$$.

The result I would like to obtain is $$-4\epsilon\csc^{2}\theta$$, which can be evaluated only on the knowledge of $${\cal X(\epsilon)},{\cal Y(\epsilon)}$$ as described above. However, I don't actually want to take limits as $$\epsilon \to 0$$ of the whole expression. I want the result to still involve terms in $$\epsilon$$ given that the above is a term in an asymptotic expansion in $$\epsilon$$. If I take $$\epsilon\to0$$ along with $$\mathcal{X}\to 0$$ and $$\mathcal{Y}\to 0$$, I will just get zero, which is no use to me.

I am stumped on how to implement this in Mathematica, even though I know exactly what calculation I am doing. Are there any elegant ways I can accomplish this?

• Maybe Series, with X[e] and Y[e] written as functions of epsilon? Oct 14, 2020 at 18:21
• Assuming X[e]~X0 e^alfa,Y[e]~Y0 e^beta as e->0,0<alfa,beta<1 (see MichaelE2's comment) gives a limit ~ 4 e^\[Beta] Y0 Csc[\[Theta]]  Oct 15, 2020 at 6:35